E.3 Radioactive decay flashcards: IB Physics SL and HL

All 75 E.3 Radioactive decay cards

1. DefinitionSL & HL

   Define an alpha particle and state its charge, mass, ionising power and penetration.

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   An alpha particle is a helium-4 nucleus, ⁴₂He, made of 2 protons and 2 neutrons. Charge +2e = +3.20 × 10⁻¹⁹ C; mass ≈ 4 u = 6.6 × 10⁻²⁷ kg. It is emitted from heavy nuclei (typically Z > 82) with a discrete kinetic energy of a few MeV. Most strongly ionising of the three radiations, therefore least penetrating: range a few cm in air, stopped by a sheet of paper or the outer layer of skin. In a magnetic field it deflects in the opposite sense to β⁻ and much less, because of its far larger mass-to-charge ratio. Exam tip: "a helium atom" loses the mark — it carries no electrons.

2. DefinitionSL & HL

   Define a beta-minus particle and state the nuclear change that produces it.

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   A β⁻ particle is a fast-moving electron, ⁰₋₁e, created and emitted at the instant a neutron in the nucleus changes into a proton: n → p + e⁻ + ν̄_e. Charge −e = −1.60 × 10⁻¹⁹ C; rest mass 9.11 × 10⁻³¹ kg ≈ u/1836. Moderately ionising and moderately penetrating: stopped by a few mm of aluminium, range up to a few m in air. Its kinetic energy takes a continuous range from zero up to a maximum. Exam tip: it is not an orbital electron — saying it "comes from the electron shells" scores zero; the electron antineutrino (not the neutrino) must be included in the equation.

3. DefinitionSL & HL

   Define a beta-plus particle and state the nuclear change that produces it.

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   A β⁺ particle is a positron, ⁰₊₁e, the antiparticle of the electron, created when a proton in the nucleus changes into a neutron: p → n + e⁺ + ν_e. Charge +e; rest mass identical to the electron, 9.11 × 10⁻³¹ kg. It is emitted by proton-rich nuclei lying below the band of stability. Like β⁻ it has a continuous energy spectrum, but it deflects the opposite way in a magnetic field. Exam tip: a free proton cannot decay this way — energy for the extra rest mass comes from the nucleus, so β⁺ emission occurs only inside a nucleus. A positron annihilates with the first electron it meets, giving two 0.511 MeV gamma photons (basis of PET scanning).

4. DefinitionSL & HL

   Define gamma radiation and state how it is produced.

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   Gamma radiation is a high-energy photon of electromagnetic radiation (typically λ < 10⁻¹² m) emitted when a daughter nucleus left in an excited state drops to a lower nuclear energy level. Zero charge, zero rest mass, speed c, undeflected by electric or magnetic fields. Weakly ionising and therefore most penetrating: intensity is reduced exponentially by absorber thickness but never completely stopped, so we quote a half-value thickness rather than a range; several cm of lead or m of concrete are needed. Exam tip: gamma emission changes neither the nucleon number A nor the proton number Z, so it never produces a new element. Discrete gamma energies are evidence for quantised nuclear energy levels.

5. DefinitionSL & HL

   Define the neutrino and antineutrino and state the experimental evidence for their existence.

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   A neutrino is a chargeless lepton of very small but non-zero mass that interacts only through the weak interaction (and gravity), so it is extremely penetrating and very hard to detect. An electron antineutrino ν̄_e accompanies β⁻ decay; an electron neutrino ν_e accompanies β⁺ decay. Evidence: the β energy spectrum is continuous, whereas a genuine two-body decay (nucleus + electron only) would give the electron a single discrete energy, exactly as the α spectrum is discrete. Pauli therefore postulated a third particle to carry away the missing energy and momentum. Exam tip: quote the α spectrum as the contrast — simply saying "energy is missing" is not enough for full credit.

6. DefinitionSL & HL

   Define the activity of a radioactive source and state its unit.

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   The activity A of a source is the number of nuclei that decay per unit time, A = ΔN/Δt. SI unit becquerel (Bq); 1 Bq = one decay per second. It is a scalar. Exam tip: activity is not the same as the measured count rate — a detector subtends only a small solid angle, is not 100 % efficient, and its reading includes background, so the corrected count rate is always smaller than the true activity but is proportional to it. Activity is proportional to the number of undecayed nuclei present, so it falls with the same half-life as N and can be used in place of N in every half-life calculation.

7. DefinitionSL & HL

   Define the half-life of a radioactive nuclide.

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   The half-life T½ is the time taken for the number of undecayed nuclei of that nuclide in a sample — or equivalently for the activity, the count rate or the mass of that nuclide — to fall to half its initial value. SI unit s (often quoted in min, days or years). Exam tip: "the time for half the sample to decay" is accepted, but "the time for the mass of the sample to halve" is not, because the daughter atoms remain and the total mass barely changes. Half-life is constant for a given nuclide and independent of the sample size, its temperature, pressure and chemical state; it follows from the constant decay probability.

8. DefinitionSL & HL

   Define ionising power and penetrating power and rank alpha, beta and gamma for each.

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   Ionising power is the number of ion pairs a radiation produces per unit length of its path in a medium; penetrating power is the thickness of material needed to absorb it. Ionising power: α ≫ β⁻ ≈ β⁺ ≫ γ, roughly 10⁴ : 10² : 1 ion pairs per mm of air. Penetration is the inverse ranking: α stopped by paper or ~5 cm of air, β by ~3 mm of aluminium, γ reduced only exponentially by cm of lead. Exam tip: the two properties are inversely related because a strongly ionising radiation loses its energy over a short path. Charge and mass explain the order: α is doubly charged and slow-moving, γ is uncharged.

9. DefinitionSL & HL

   Define the mass defect of a nucleus.

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   The mass defect Δm of a nucleus is the difference between the total mass of its constituent nucleons when completely separated and the actual mass of the assembled nucleus: Δm = Zm_p + (A − Z)m_n − m_nucleus. It is always positive and is normally quoted in u or in MeV c⁻². Exam tip: two marks are commonly lost — using the mass of the neutral atom without allowing for the Z electron masses, and subtracting the wrong way round to give a negative value. The missing mass appeared as the energy released when the nucleus formed; mass–energy overall is conserved, so "mass is lost" must always be qualified by "converted to energy".

10. DefinitionSL & HL

    Define the binding energy of a nucleus.

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    The binding energy of a nucleus is the minimum energy required to separate the nucleus completely into its individual protons and neutrons, with all of them at rest and infinitely far apart; equivalently, it is the energy released when those nucleons come together to form the nucleus. It equals Δmc² and is normally quoted in MeV. Exam tip: it is not energy stored inside the nucleus — that is the classic wrong answer; a nucleus with a large binding energy is more tightly bound and harder to break apart, not less stable. The phrases "completely separated" and "into its constituent nucleons" carry the marks; "to break up the nucleus" alone is too vague.

11. DefinitionSL & HL

    Define binding energy per nucleon and state why it, rather than total binding energy, measures nuclear stability.

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    Binding energy per nucleon is the total binding energy of a nucleus divided by its nucleon number A, quoted in MeV per nucleon. It is the average energy needed to remove one nucleon from the nucleus. It is the correct measure of stability because total binding energy simply grows with the number of nucleons, so a large unstable nucleus such as U-235 (about 1780 MeV) exceeds a very stable one such as Fe-56 (about 492 MeV). Per nucleon, Fe-56 has the greater value (≈ 8.8 MeV compared with ≈ 7.6 MeV) and is the more stable. Exam tip: always state the units "MeV per nucleon" and say "higher value means more stable".

12. DefinitionSL & HL

    Define the unified atomic mass unit and the unit MeV c⁻².

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    The unified atomic mass unit u is one twelfth of the mass of an unbound, neutral atom of carbon-12 in its ground state: 1 u = 1.66 × 10⁻²⁷ kg. Multiplying by c² and converting to electronvolts gives the mass–energy equivalence 1 u = 931.5 MeV c⁻², so a mass defect of 1 u corresponds to 931.5 MeV of energy released. MeV c⁻² is therefore a unit of mass, and MeV a unit of energy. Exam tip: the commonest slip is quoting a mass in MeV or an energy in MeV c⁻²; check the c⁻². Useful reference values: m_p = 1.007276 u, m_n = 1.008665 u, m_e = 0.000549 u.

13. DefinitionSL & HL

    Define the band (curve) of stability on a neutron number against proton number graph.

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    The band of stability is the narrow region on an N against Z plot occupied by the stable nuclides. For light nuclei up to about Z = 20 it follows the line N = Z; above this it curves upwards so that heavier stable nuclei have progressively more neutrons than protons (N/Z rises to about 1.5 for lead), because extra neutrons add strong-force attraction without adding electrostatic repulsion. No nuclide with Z > 83 is stable. Exam tip: nuclides above the band (neutron rich) decay by β⁻; nuclides below the band (proton rich) decay by β⁺ or electron capture; nuclides beyond Z = 83 decay mainly by α, moving diagonally down towards the band.

14. EquationSL & HLData booklet: Yes

    State the mass–energy equivalence equation used in nuclear calculations and define every symbol.

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    ΔE = Δmc². ΔE = energy released or required (J, or MeV if masses are in MeV c⁻²); Δm = mass defect or change in mass (kg, or u); c = speed of light in vacuum, 3.00 × 10⁸ m s⁻¹. Data-booklet status: printed. Valid for any process in which rest mass changes: nuclear decay, fission, fusion, annihilation. Shortcut: energy in MeV = Δm in u × 931.5. Common misuse: mixing units — putting Δm in u while c is in m s⁻¹ gives an answer 10²⁷ times too small; convert with 1 u = 1.66 × 10⁻²⁷ kg first. Sanity check: Δm = 0.0186 u gives 0.0186 × 931.5 ≈ 17.3 MeV.

15. EquationSL & HLData booklet: No – memorise

    State the equation for the mass defect of a nuclide ᴬ_Z X and define every symbol.

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    Δm = Zm_p + (A − Z)m_n − m_nucleus. Δm = mass defect (u or kg); Z = proton number; A = nucleon number; A − Z = neutron number N; m_p = 1.007276 u, m_n = 1.008665 u, m_nucleus = measured nuclear mass (u). Data-booklet status: not printed — memorise. Binding energy follows as E_b = Δmc² = Δm(u) × 931.5 MeV. Common misuse: using atomic (not nuclear) masses without subtracting Zm_e — for helium-4, atomic mass 4.002602 u contains two electrons. Sanity check for ⁴₂He: 2(1.007276) + 2(1.008665) − 4.001506 = 0.030376 u, giving 28.3 MeV, that is 7.07 MeV per nucleon.

16. EquationSL & HLData booklet: Yes

    State the mass–energy conversion factor for the unified atomic mass unit and show how it is used.

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    1 u = 931.5 MeV c⁻², equivalently 1 u = 1.66 × 10⁻²⁷ kg and 1 eV = 1.60 × 10⁻¹⁹ J. Symbols: u = unified atomic mass unit; MeV c⁻² = mass unit; MeV = 1.60 × 10⁻¹³ J. Data-booklet status: the conversions and constants are printed. Use: energy released in MeV = mass defect in u × 931.5, avoiding any conversion to kg and joules. Common misuse: writing 1 u = 931.5 MeV (missing c⁻²), which equates a mass to an energy. Sanity check: the electron rest mass 0.000549 u × 931.5 = 0.511 MeV c⁻², the familiar annihilation photon energy.

17. EquationSL & HLData booklet: No – derive

    State how binding energy per nucleon is calculated from a nuclear mass.

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    E_b/A = Δmc²/A, with Δm = Zm_p + (A − Z)m_n − m_nucleus. E_b = total binding energy (MeV or J); A = nucleon number (no unit); E_b/A = binding energy per nucleon (MeV per nucleon). Data-booklet status: derived from the printed ΔE = Δmc²; the division by A must be remembered. Energy released in any nuclear reaction = A_products × (E_b/A)_products − A_reactants × (E_b/A)_reactants, or more simply the total mass difference × 931.5. Common misuse: dividing by the number of protons, or forgetting to divide at all when comparing stability. Sanity check: Fe-56 has E_b ≈ 492 MeV, so E_b/A ≈ 8.8 MeV per nucleon, the maximum of the curve.

18. EquationSL & HLData booklet: No – memorise

    State the general equation for alpha decay and give a worked example.

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    ᴬ_Z X → ᴬ⁻⁴_Z₋₂ Y + ⁴₂He (often written ⁴₂α). Symbols: A = nucleon number, Z = proton number, X = parent nuclide, Y = daughter nuclide. Data-booklet status: not printed — memorise the rule A decreases by 4 and Z by 2. Example: ²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He. Both nucleon number and proton number (charge) must balance on each side. Common misuse: writing the alpha as an atom or forgetting that the daughter is a different element, and quoting mass number 4 with charge zero. Sanity check: 226 = 222 + 4 and 88 = 86 + 2. Energy released appears mainly as the discrete kinetic energy of the alpha particle.

19. EquationSL & HLData booklet: No – memorise

    State the general equation for beta-minus decay, including the underlying nucleon change.

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    ᴬ_Z X → ᴬ_Z₊₁ Y + ⁰₋₁e + ν̄_e, underpinned by n → p + e⁻ + ν̄_e. Symbols: A = nucleon number (unchanged); Z = proton number (increases by one); ⁰₋₁e = electron; ν̄_e = electron antineutrino (charge 0, lepton number −1). Data-booklet status: not printed — memorise. Example: ¹⁴₆C → ¹⁴₇N + ⁰₋₁e + ν̄_e. Occurs for neutron-rich nuclei above the band of stability. Common misuse: omitting the antineutrino, or writing ν instead of ν̄; both lose the mark because lepton number must balance (0 = 0 + 1 − 1). Sanity check: charge balance 6 = 7 + (−1); nucleon number 14 = 14.

20. EquationSL & HLData booklet: No – memorise

    State the general equation for beta-plus decay, including the underlying nucleon change.

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    ᴬ_Z X → ᴬ_Z₋₁ Y + ⁰₊₁e + ν_e, underpinned by p → n + e⁺ + ν_e. Symbols: A = nucleon number (unchanged); Z = proton number (decreases by one); ⁰₊₁e = positron; ν_e = electron neutrino (lepton number +1). Data-booklet status: not printed — memorise. Example: ²²₁₁Na → ²²₁₀Ne + ⁰₊₁e + ν_e. Occurs for proton-rich nuclei below the band of stability. Common misuse: pairing the positron with an antineutrino — lepton number would then be 0 = −1 − 1, which fails; the positron has lepton number −1, so a neutrino (+1) is required. Sanity check: charge 11 = 10 + 1; nucleon number 22 = 22.

21. EquationSL & HLData booklet: No – derive

    State the half-life form of the decay law used at SL and define every symbol.

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    N = N₀(½)^(t/T½), and identically A = A₀(½)^(t/T½) and m = m₀(½)^(t/T½). N = number of undecayed nuclei remaining after time t; N₀ = initial number; t = elapsed time (s); T½ = half-life (same unit as t); t/T½ = number of half-lives n. Data-booklet status: not printed at SL — memorise or derive by repeated halving. Fraction remaining after n half-lives = (½)ⁿ; fraction decayed = 1 − (½)ⁿ. Common misuse: reading "three quarters have decayed" as three half-lives instead of two. Sanity check: after 4 half-lives 1/16 = 6.25 % remains, so an activity of 800 Bq falls to 50 Bq.

22. Graph/diagramSL & HL

    Describe the neutron number against proton number graph for stable nuclides and explain what it shows.

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    Axes: neutron number N (no unit) on the y-axis, proton number Z (no unit) on the x-axis, both plotted as integers. The stable nuclides form a narrow band that lies on the line N = Z up to about Z = 20, then curves increasingly above it, reaching N/Z ≈ 1.5 near lead, and stops at Z = 83. The dashed reference line N = Z is drawn to show the departure. Interpretation: extra neutrons supply extra short-range strong-force attraction without adding Coulomb repulsion, so heavier stable nuclei need a neutron excess. Reading it: a point above the band decays by β⁻ (moving down-right, N falls by 1, Z rises by 1); below the band, β⁺ (up-left); beyond Z = 83, α decay moves the point down 2 and left 2.

23. Graph/diagramSL & HL

    Describe the binding energy per nucleon against nucleon number curve and state what each feature means.

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    Axes: binding energy per nucleon E_b/A in MeV per nucleon on the y-axis, nucleon number A on the x-axis. The curve rises steeply and unevenly for light nuclei (with peaks at ⁴He, ¹²C, ¹⁶O), reaches a maximum of about 8.8 MeV per nucleon near A = 56 (Fe-56, the most stable nuclide), then falls slowly to about 7.6 MeV per nucleon at U-235. Higher on the curve means more tightly bound. Fusion of nuclei with A < 56 and fission of nuclei with A > 56 both move products up the curve, so both release energy. Energy released = (final E_b/A × A summed) − (initial), or equivalently the increase in total binding energy. Some booklets plot −E_b/A, giving an inverted well.

24. Graph/diagramSL & HL

    Describe the shape of an activity against time decay curve and explain how to obtain the half-life from it.

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    Axes: activity A/Bq (or corrected count rate in s⁻¹) on the y-axis, time t on the x-axis. The curve starts at A₀, falls steeply, and flattens asymptotically towards zero without ever reaching it — it is exponential, so it has a constant ratio property. Method: read the time for the activity to fall from A₀ to A₀/2, then from A₀/2 to A₀/4, then A₀/4 to A₀/8; each interval is one half-life. Averaging several such intervals reduces random error. The gradient at any point equals −λA, so the curve is steepest at t = 0. Increasing the initial sample size raises the whole curve without changing its shape or the half-life; a shorter half-life makes the curve fall more steeply.

25. Graph/diagramSL & HL

    Compare the energy spectrum of alpha particles with that of beta particles and state what the difference shows.

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    Axes: number of particles emitted per unit energy interval on the y-axis, kinetic energy E_k/MeV on the x-axis. Alpha: one (or a few) very narrow, tall spikes at discrete energies of a few MeV, because the decay is a two-body process in which energy sharing is fixed by conservation of momentum, and because nuclear energy levels are quantised. Beta: a continuous, smooth distribution from zero rising to a broad peak and falling to zero at a definite maximum energy E_max, which equals the total decay energy. Significance: the continuous beta spectrum shows a third particle, the (anti)neutrino, shares the energy and momentum; without it, energy and momentum conservation would appear to be violated.

26. Graph/diagramSL & HL

    Describe how the count rate varies with absorber thickness for alpha, beta and gamma, and explain the shapes.

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    Axes: corrected count rate in s⁻¹ on the y-axis, absorber thickness x in mm on the x-axis, with background already subtracted. Alpha: an almost flat line that drops abruptly to zero at a definite range (a few cm of air, or paper) — alphas lose energy steadily and all stop at nearly the same depth. Beta: falls steeply and roughly exponentially, reaching zero at a few mm of aluminium, with a spread of ranges because the energies are continuous. Gamma: a true exponential decrease, I = I₀e^(−μx), that never reaches zero, characterised by a half-value thickness x½ = ln2/μ read where the count rate halves. A denser absorber increases μ and steepens the gamma curve.

27. Graph/diagramSL & HL

    Describe the deflection of alpha, beta-minus, beta-plus and gamma radiation in a magnetic field and in an electric field.

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    In a magnetic field directed into the page with the beam travelling to the right: α (charge +2e) curves one way, β⁻ (−e) curves the opposite way, β⁺ (+e) curves the same way as α, and γ is undeflected. Radius r = mv/(qB), so the deflection is far greater for β because m/q is about 3600 times smaller — beta paths curve tightly while alpha paths bend only slightly. In a uniform electric field between parallel plates the tracks are parabolic: α and β⁺ bend towards the negative plate, β⁻ towards the positive plate, γ passes straight through. Sketching tip: always label the field direction, mark the charges, and show γ as an undeviated straight line.

28. Concept/explainSL & HL

    Compare alpha, beta and gamma radiation in terms of their nature, charge, mass, ionising power and penetration.

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    • α is a helium-4 nucleus ⁴₂He, charge +2e, mass ≈ 4 u, emitted with discrete energies of a few MeV
    • β⁻ is a fast electron created in the nucleus, charge −e, mass ≈ 1/1836 u, emitted with a continuous range of energies; β⁺ is a positron, charge +e
    • γ is a high-energy photon, zero charge and zero rest mass, travelling at c
    • ionising power falls α ≫ β ≫ γ because ionisation depends on charge and on the time spent near atoms, and α is doubly charged and slow
    • penetration is the reverse order: α is stopped by a few cm of air or a sheet of paper, β by a few mm of aluminium, γ is only attenuated by several cm of lead. Exam tip: γ is reduced, never completely stopped — write "attenuated", not "absorbed by lead".
29. Concept/explainSL & HL

    Explain how α, β and γ radiation behave when passed through a uniform magnetic field and through a uniform electric field.

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    • Only charged radiation is deflected, so γ passes through both fields undeviated
    • in a magnetic field the force is F = qvB and the path is circular with radius r = mv/(qB), so the deflection depends on the charge-to-mass ratio
    • α (+2e, 4 u) and β⁻ (−e, 1/1836 u) curve in OPPOSITE senses
    • β is deflected far more than α because its mass is about 7000 times smaller, so its q/m is far larger
    • in an electric field between parallel plates α is attracted towards the negative plate and β⁻ towards the positive plate, both following a parabolic path
    • β⁺ deflects the same way as α but much more strongly. Exam tip: the commonest error is drawing α deflected more than β — the massive α barely bends.
30. Concept/explainSL & HL

    Explain what happens inside the nucleus during β⁻ and β⁺ decay and write the full decay equations.

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    • β⁻ decay: a neutron becomes a proton, n → p + e⁻ + ν̄_e
    • Z increases by 1 while A is unchanged, e.g. ¹⁴₆C → ¹⁴₇N + e⁻ + ν̄_e
    • β⁺ decay: a proton becomes a neutron, p → n + e⁺ + ν_e, so Z decreases by 1
    • the electron or positron does not pre-exist in the nucleus — it is created at the instant of decay
    • both are weak-interaction processes
    • charge, nucleon (baryon) number and lepton number are all conserved: L = +1 for e⁻ and ν_e, L = −1 for e⁺ and ν̄_e. Exam tip: the ANTIneutrino accompanies the electron and the neutrino accompanies the positron — reversing them is the most frequently lost mark.
31. Concept/explainSL & HL

    Explain how the energy spectra of α and β decay provided evidence for the existence of the neutrino.

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    • α decay is a two-body decay, so conservation of energy and momentum forces the α to carry a fixed share of the energy released: the α spectrum is DISCRETE, a set of sharp lines
    • the β⁻ spectrum is CONTINUOUS, running from zero up to a maximum energy E_max
    • if β decay were also two-body the electrons would all have the same energy, so energy and momentum appeared not to be conserved, and the observed angular momentum was also wrong
    • Pauli (1930) proposed a third, neutral, almost massless particle that shares the energy — later named the neutrino
    • E_max equals the total energy released, corresponding to the antineutrino carrying essentially none
    • direct detection followed in 1956. NOS: physicists preferred to predict an undetected particle rather than abandon conservation of energy.
32. Concept/explainSL & HL

    Explain how the graph of neutron number against proton number predicts which decay mode an unstable nuclide will undergo.

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    • The chart plots neutron number N on the y-axis against proton number Z on the x-axis; stable nuclides lie in a narrow band of stability
    • for light nuclides the band follows N ≈ Z
    • for heavier nuclides it curves ABOVE the N = Z line because extra neutrons supply strong-force attraction without adding electrostatic repulsion
    • a nuclide ABOVE the band is neutron-rich and decays by β⁻ (n → p), moving one step down and one step right, towards the band
    • a nuclide BELOW the band is proton-rich and decays by β⁺ or electron capture, moving one step up and one step left
    • nuclides with Z > 83 are too large for the short-range strong force to bind and decay by α, moving 2 down and 2 left. Exam tip: state the direction of movement on the chart, not just the name of the decay.
33. Concept/explainSL & HL

    Outline the main sources of background radiation and explain why background must be measured in every radioactivity experiment.

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    • Background radiation is the ionising radiation always present from sources other than the one under investigation
    • natural sources: radon gas seeping from rocks and soil, which is the largest contribution in most places; cosmic rays; rocks and building materials such as granite; food, drink and the potassium-40 inside our own bodies
    • artificial sources: medical X-rays and nuclear medicine, weapons-test fallout and the nuclear industry
    • background varies with location, altitude and local geology, so it must be measured, not assumed
    • procedure: record the count with no source present, over a long time to reduce the fractional uncertainty, and subtract it from every reading to give the corrected count rate
    • ignoring it makes measured half-lives too long, because the tail of the curve flattens onto the background level.
34. Concept/explainSL & HL

    Explain what is meant by saying that radioactive decay is both random and spontaneous, and state the experimental evidence.

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    • Spontaneous means the decay is not triggered by anything external: the rate is unaffected by temperature, pressure, chemical bonding or physical state, because it is a nuclear rather than an electronic process
    • random means it is impossible to predict which nucleus will decay next, or when any particular nucleus will decay
    • every undecayed nucleus has the same constant probability of decaying in the next unit of time, independent of its age and of what its neighbours do
    • these two statements together force the number remaining to fall exponentially and give a constant half-life
    • evidence: the count rate from a GM tube fluctuates about a mean value even for a source of very long half-life
    • the decay law is statistical and is only reliable for large numbers of nuclei. Exam tip: random and spontaneous are separate ideas — define both for two marks.
35. Concept/explainSL & HLData booklet: No – memorise

    Define activity and half-life, and explain why the half-life of a given isotope is a constant.

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    • Activity A is the number of nuclei that decay per unit time; the SI unit is the becquerel (Bq), where 1 Bq = 1 decay per second
    • half-life T½ is the time taken for the number of undecayed nuclei of an isotope in a sample — and therefore also for the activity, or for the mass of that isotope — to fall to half its initial value
    • because each nucleus has a fixed decay probability, the decrease is exponential, so the time to halve is the same from ANY starting point on the curve
    • after n half-lives the fraction remaining is (½)ⁿ
    • the measured count rate is proportional to activity but always smaller, because of detector geometry (solid angle), detector efficiency and absorption in the air and window. Exam tip: half-life is not the time for a source to "half decay and then stop", and it is not half the total lifetime.
36. Concept/explainSL & HLData booklet: Yes

    Explain the meaning of mass defect and nuclear binding energy and how they are related.

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    • The mass of a nucleus is always LESS than the total mass of the individual protons and neutrons that make it up; that difference is the mass defect Δm
    • binding energy is the energy that must be supplied to separate a nucleus completely into its individual nucleons, at rest and far apart
    • equivalently it is the energy released when free nucleons come together to form the nucleus
    • the two are linked by Einstein's mass–energy equivalence E = Δmc² (booklet)
    • working in nuclear units, 1 u is equivalent to 931.5 MeV, so binding energy in MeV = Δm in u × 931.5
    • binding energy per nucleon (total ÷ A) is the fair measure of stability, since total binding energy simply grows with size. Exam tip: binding energy is not energy stored in the nucleus — the bound nucleus sits at a LOWER energy than its separated parts.
37. Concept/explainSL & HL

    Explain the shape of the binding energy per nucleon curve and use it to show why both fusion and fission release energy.

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    • The curve plots binding energy per nucleon in MeV (y) against nucleon number A (x)
    • it rises steeply for light nuclei, peaks at about 8.8 MeV near A = 56 (iron-56, the most stable nuclide), then falls slowly to about 7.6 MeV at uranium
    • ⁴He, ¹²C and ¹⁶O sit noticeably above the local trend because their nucleons are especially tightly packed
    • a nuclear reaction releases energy when the PRODUCTS have a higher binding energy per nucleon than the reactants, since the products are more tightly bound and the surplus mass appears as kinetic energy and photons
    • so fusing nuclei with A < 56 releases energy, and splitting nuclei with A > 56 releases energy
    • energy released ≈ A × (increase in binding energy per nucleon). Exam tip: this is not a graph of TOTAL binding energy, which increases all the way to uranium.
38. Concept/explainSL & HL

    Describe the properties of the strong nuclear force and explain why there are no stable nuclides beyond Z = 83.

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    • The strong nuclear force acts attractively between every pair of nucleons — p–p, p–n and n–n alike — so it is charge-independent
    • it is very short range, falling effectively to zero beyond about 3 fm, and becomes strongly REPULSIVE below about 0.5 fm, which fixes the spacing of nucleons and gives all nuclei nearly the same density
    • within its range it is roughly 100 times stronger than the electrostatic repulsion between two protons
    • because it is short range, each nucleon only attracts its nearest neighbours, so the attraction saturates, whereas electrostatic repulsion is long range and acts between EVERY pair of protons
    • as Z rises the repulsion therefore grows faster than the attraction; extra neutrons help, but beyond Z = 83 no ratio works and every nuclide is unstable. Exam tip: you must say both "short range" and "acts between all nucleons".
39. Concept/explainSL & HL

    Explain the principle of carbon-14 dating and state the assumptions and limitations of the method.

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    • Cosmic-ray neutrons continually create ¹⁴C in the upper atmosphere; it mixes as CO₂ and enters the food chain
    • while an organism lives it exchanges carbon with its surroundings, so its ¹⁴C to ¹²C ratio, and hence its activity per gram of carbon, stays equal to the atmospheric value
    • at death the exchange stops and the trapped ¹⁴C decays by β⁻ with T½ = 5730 years, so the specific activity falls exponentially
    • comparing the sample's activity per gram with that of living material gives the number of half-lives elapsed, and hence the age
    • assumptions: the atmospheric ¹⁴C concentration was the same in the past, and the sample has not been contaminated with modern carbon
    • limit ≈ 50 000 years (about 9 half-lives) before the count rate is lost in the background
    • older rocks are dated with long-lived pairs such as U-238/Pb-206 or K-40/Ar-40.
40. Concept/explainSL & HL

    Explain why gamma rays are emitted after alpha or beta decay and what this reveals about the nucleus.

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    • After emitting an α or β particle the daughter nucleus is usually left in an EXCITED state, with its nucleons in a higher energy configuration
    • it de-excites by emitting a γ-ray photon of energy E = hf = ΔE, the difference between two nuclear energy levels
    • neither A nor Z changes, so the nuclide is unchanged and γ emission must not be balanced in the A/Z arithmetic
    • the measured γ energies are DISCRETE, typically 0.1–3 MeV, which is direct evidence that the nucleus has quantised energy levels, exactly as atomic line spectra evidence quantised electron levels
    • de-excitation is normally almost instantaneous, but metastable states such as Tc-99m survive for hours and are used as medical tracers
    • being uncharged, γ is weakly ionising and highly penetrating. Exam tip: γ emission is a change of state, not a change of element.
41. Worked problemSL & HLData booklet: No – memorise

    Uranium-238 decays by alpha emission; the daughter then decays by beta-minus twice in succession. Determine the final nuclide and write the three equations.

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    Rule: α emission reduces A by 4 and Z by 2; β⁻ leaves A unchanged and increases Z by 1. Step 1: ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He. Step 2: ²³⁴₉₀Th → ²³⁴₉₁Pa + e⁻ + ν̄_e. Step 3: ²³⁴₉₁Pa → ²³⁴₉₂U + e⁻ + ν̄_e. Final nuclide: ²³⁴₉₂U — uranium again, but four nucleons lighter than the parent. Overall change: ΔA = −4, ΔZ = 0. Check/Trap: nucleon number and proton number must each balance on both sides of every line; the antineutrino carries no charge and no nucleons so it does not affect the arithmetic, but omitting it loses a mark in a "write the full equation" question.

42. Worked problemSL & HLData booklet: No – memorise

    A radioactive source has an initial activity of 480 Bq and a half-life of 15 minutes. Calculate its activity after 1.0 hour.

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    Principle: activity halves every half-life, A = A₀ × (½)ⁿ where n is the number of half-lives elapsed. Number of half-lives: n = t/T½ = 60 min / 15 min = 4.0. Fraction remaining: (½)⁴ = 1/16 = 0.0625. Activity: A = 480 Bq × 1/16 = 30 Bq. Answer: 30 Bq (2 s.f.). Check/Trap: work through the halvings as a check — 480 → 240 → 120 → 60 → 30 Bq. The classic Paper 1 error is dividing by 4 instead of halving 4 times, giving 120 Bq. Also make sure the time and the half-life are in the SAME unit before dividing.

43. Worked problemSL & HLData booklet: No – memorise

    A sample contains 8.0 × 10²⁰ undecayed nuclei of an isotope with a half-life of 12 days. Determine the number remaining, and the number that have decayed, after 36 days.

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    Principle: N = N₀ × (½)ⁿ, n = t/T½. Number of half-lives: n = 36 days / 12 days = 3.0. Fraction remaining: (½)³ = 1/8 = 0.125. Nuclei remaining: N = 8.0 × 10²⁰ × 0.125 = 1.0 × 10²⁰. Nuclei that have DECAYED: N₀ − N = 8.0 × 10²⁰ − 1.0 × 10²⁰ = 7.0 × 10²⁰. Answers: 1.0 × 10²⁰ remaining, 7.0 × 10²⁰ decayed (2 s.f.). Check/Trap: read the question wording carefully — "remaining", "decayed" and "become the daughter product" are three different answers from the same calculation, and the number decayed is the one most often given wrongly.

44. Worked problemSL & HLData booklet: Yes

    The atomic mass of lithium-7 is 7.01600 u. Determine its mass defect, its total binding energy in MeV and its binding energy per nucleon. Take m(¹H) = 1.00783 u and m(n) = 1.00867 u.

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    Composition: ⁷₃Li has Z = 3 protons and A − Z = 4 neutrons; using atomic masses, take 3 hydrogen atoms plus 4 neutrons so the 3 electrons cancel. Total mass of parts: 3 × 1.00783 = 3.02349 u; 4 × 1.00867 = 4.03468 u; sum = 7.05817 u. Mass defect: Δm = 7.05817 − 7.01600 = 0.04217 u. Binding energy: E = Δm × 931.5 MeV u⁻¹ = 0.04217 × 931.5 = 39.3 MeV. Per nucleon: 39.3 / 7 = 5.61 MeV per nucleon. Answers: Δm = 0.04217 u, E = 39.3 MeV, 5.61 MeV per nucleon (3 s.f.). Check/Trap: 5.6 MeV is sensibly below the 8.8 MeV peak for a light nucleus. Keep 5 decimal places throughout — Δm is a small difference of large numbers, so early rounding destroys the answer.

45. Worked problemSL & HLData booklet: Yes

    Radium-226 (225.97710 u nuclear mass) decays to radon-222 (221.97036 u) by alpha emission; the alpha particle has mass 4.00151 u. Determine the energy released and the kinetic energy of the alpha particle.

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    Mass difference: Δm = 225.97710 − (221.97036 + 4.00151) = 225.97710 − 225.97187 = 0.00523 u. Energy released: Q = Δm × 931.5 = 0.00523 × 931.5 = 4.87 MeV. This appears as kinetic energy shared between the α and the recoiling radon nucleus. Momentum conservation (parent at rest): m_α v_α = M_Rn v_Rn, and since KE = p²/2m with equal magnitudes of p, KE_α / Q = M_Rn / (M_Rn + m_α) = 222/226 = 0.982. So KE_α = 0.982 × 4.87 = 4.78 MeV. Answers: 4.87 MeV released, 4.78 MeV to the α (3 s.f.). Check/Trap: the α takes almost all the energy but NOT all of it; forgetting the recoil, or assuming the two share equally, are both common. Momentum is shared equally in magnitude, kinetic energy is not.

46. Worked problemSL & HLData booklet: No – memorise

    A piece of charcoal from an archaeological site has a carbon-14 activity per gram of carbon that is one eighth of that of living wood. The half-life of carbon-14 is 5730 years. Determine the age of the charcoal.

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    Principle: after death the ¹⁴C activity falls exponentially, A = A₀ × (½)ⁿ. Fraction remaining: A/A₀ = 1/8 = (½)³, so n = 3 half-lives. Age: t = n × T½ = 3 × 5730 = 17190 years ≈ 1.7 × 10⁴ years (2 s.f.). Assumptions used: the ¹⁴C fraction in the atmosphere was the same when the tree died, and the sample has not been contaminated. Check/Trap: the comparison must be activity PER GRAM OF CARBON, so the two samples need not have the same mass — quoting a raw count rate ratio for samples of different mass is a standard trap. Also confirm the background count has been subtracted before the ratio is taken; if not, the calculated age comes out too small.

47. Worked problemSL & HLData booklet: No – derive

    A nucleus with A = 236 and binding energy per nucleon 7.6 MeV splits into two fragments each with A = 118 and binding energy per nucleon 8.5 MeV. Determine the energy released, in MeV and in joules.

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    Principle: energy released = total binding energy of products − total binding energy of reactant, because the products are more tightly bound. Reactant: BE = 236 × 7.6 = 1794 MeV. Products: BE = 2 × (118 × 8.5) = 2 × 1003 = 2006 MeV. Energy released: ΔE = 2006 − 1794 = 212 MeV ≈ 2.1 × 10² MeV (2 s.f.). Alternatively ΔE = A × Δ(BE per nucleon) = 236 × (8.5 − 7.6) = 212 MeV. In joules: 212 × 10⁶ × 1.60 × 10⁻¹⁹ = 3.4 × 10⁻¹¹ J. Check/Trap: the sign convention — energy is released because binding energy per nucleon INCREASES; students often subtract the wrong way and report a negative value. Nucleon number is conserved (236 = 2 × 118), so the shortcut form is valid here.

48. Worked problemSL & HLData booklet: Yes

    A cobalt-60 nucleus de-excites by emitting a gamma photon of energy 1.33 MeV. Determine the frequency and wavelength of the photon.

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    Convert to joules: E = 1.33 × 10⁶ × 1.60 × 10⁻¹⁹ = 2.13 × 10⁻¹³ J. Frequency from E = hf: f = E/h = 2.13 × 10⁻¹³ / 6.63 × 10⁻³⁴ = 3.21 × 10²⁰ Hz. Wavelength from c = fλ (or λ = hc/E): λ = 3.00 × 10⁸ / 3.21 × 10²⁰ = 9.35 × 10⁻¹³ m. Answers: f = 3.21 × 10²⁰ Hz, λ = 9.35 × 10⁻¹³ m (3 s.f.). Check/Trap: a wavelength of about 10⁻¹² m is smaller than an atom and comparable with nuclear dimensions — exactly what is expected for a γ-ray, so the answer is sensible. The trap is putting 1.33 straight into E = hf without converting MeV to joules, which gives a nonsense answer 10⁶ times too small.

49. Exam technique/trapSL & HL

    Explain the command-term expectations and the standard misconceptions in questions on half-life and radioactive decay.

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    Command terms: "state" the half-life needs only a value and unit; "define" needs the full sentence — the time for the number of undecayed nuclei, the activity, or the mass of the isotope to fall to half its initial value; "explain" why decay is exponential needs the constant-probability argument, not a description of the graph; "determine" allows any valid route but needs working and a unit; "suggest" invites a reasoned possibility and is marked generously. Misconceptions to avoid: half-life is not half the time for the sample to disappear; the source does not stop after two half-lives; half-life is not affected by temperature, pressure or chemical state; the daughter nuclide may itself be radioactive, so the measured count rate need not fall to background. Always subtract background before quoting a count rate, and state which quantity has halved.

50. Exam technique/trapSL & HLData booklet: Yes

    Explain the unit and conversion traps in mass defect and binding energy calculations.

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    Trap 1: mixing kilograms and unified mass units. Either work in u and multiply by 931.5 MeV u⁻¹, or convert with 1 u = 1.66 × 10⁻²⁷ kg and use E = Δmc² in joules — never mix the two routes. Trap 2: rounding early. Δm is a small difference between large masses, so keep 5 decimal places until the subtraction is done; rounding the nucleon masses to 3 s.f. can change the answer by 50%. Trap 3: atomic versus nuclear masses. If the data are ATOMIC masses, build the parts from hydrogen atoms plus neutrons so the electron masses cancel; if they are NUCLEAR masses, use proton and neutron masses. Trap 4: MeV c⁻² is a mass unit and MeV is an energy unit — check the exponent of your answer, since binding energies per nucleon must come out between about 1 and 9 MeV.

51. Exam technique/trapSL & HL

    Describe an experiment to identify the type or types of radiation emitted by an unknown sealed source, including variables, safety and limitations.

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    Apparatus: sealed source in a holder, GM tube and scaler/ratemeter, absorbers (sheet of paper, 5 mm aluminium, 2–3 cm lead), metre rule, stopwatch, tongs. Method: first record the background count for at least 5 minutes with no source present and find the mean background rate. Place the source at a fixed distance (a few cm) from the window, record the count over a fixed interval with no absorber, then with paper, then with aluminium, then with lead; repeat each three times and subtract background. Interpretation: a large drop with paper indicates α; a further large drop with aluminium indicates β; a count that survives lead indicates γ. Controlled variables: source-to-detector distance, counting time, tube voltage, geometry. Safety: handle with tongs, keep the source at arm's length, point away from the body, minimise exposure time, return to the lead-lined store. Limitations: counting is random so long counts are needed; a weak β component can be masked by a strong γ.

52. Exam technique/trapSL & HL

    Explain how counting statistics affect radioactivity measurements, and distinguish random from systematic error in this context.

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    Because decay is random, a repeated count of N registered events has an inherent (Poisson) uncertainty of about √N, so the FRACTIONAL uncertainty is √N/N = 1/√N. Consequence: counting 100 events gives ±10%, but counting 10 000 events gives ±1% — the way to improve precision is to count for longer or use a stronger source, not to repeat a short count and average sloppily. Random errors here also come from fluctuations in background. Systematic errors come from a fixed cause: not subtracting background at all (raises every reading), a moving source-to-detector distance, dead time at high count rates (lowers high readings only, so it also distorts the shape of a decay curve), and a mis-set tube voltage. Precision versus accuracy: a long count is precise but still inaccurate if background was never subtracted. Show error bars of ±√N on plotted counts.

53. DefinitionHL only

    Define the decay constant and give its probability interpretation.

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    The decay constant λ is the probability per unit time that a given undecayed nucleus will decay, equal to the fraction of the nuclei present that decay per unit time. SI unit s⁻¹ (also min⁻¹, y⁻¹). It is a scalar and is a constant characteristic of the nuclide, independent of sample size, temperature or chemical state. Exam tip: the mark requires "per unit time" — "the probability that a nucleus decays" alone is dimensionally meaningless. A large λ means a short half-life, since T½ = ln2/λ. From λ = −(1/N)(dN/dt) it follows that A = λN, which is why activity is proportional to the number of undecayed nuclei remaining.

54. DefinitionHL only

    Define corrected count rate and explain why it, rather than the raw reading, must be used.

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    The corrected count rate is the measured count rate from a detector minus the mean background count rate measured under the same conditions with the source removed. Unit s⁻¹ (or counts per minute). Exam tip: without correction the decay curve appears to flatten out at a non-zero level, so the apparent half-life grows steadily as the sample decays and a ln(count rate) against t plot curves upwards instead of being straight — a classic Paper 3 fault to identify. The corrected count rate is proportional to, but much smaller than, the true activity because the Geiger tube subtends a small solid angle, is less than 100 % efficient, and some radiation is absorbed by air and the tube window.

55. EquationHL onlyData booklet: Yes

    State the exponential decay law for the number of undecayed nuclei and define every symbol.

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    N = N₀e^(−λt). N = number of undecayed nuclei remaining at time t; N₀ = number present at t = 0; λ = decay constant (s⁻¹); t = time (s); e = base of natural logarithms. Data-booklet status: printed. Conditions: λ and t must use the same time unit, and the sample must be large enough for the statistical law to hold. Rearranged forms to know: t = (1/λ)ln(N₀/N) and ln N = ln N₀ − λt. Common misuse: dropping the minus sign, or substituting a half-life for 1/λ. Sanity check: at t = T½, λt = ln2 = 0.693 and e^(−0.693) = 0.500, so exactly half remains.

56. EquationHL onlyData booklet: Yes

    State the relationship between activity and the number of undecayed nuclei and define every symbol.

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    A = λN. A = activity (Bq, that is s⁻¹); λ = decay constant (s⁻¹); N = number of undecayed nuclei present. Data-booklet status: printed. It follows from the definition of λ as the decay probability per unit time. Conditions: λ in s⁻¹ if A is in Bq; N must be the number of nuclei, not the mass or the number of moles, so use N = (m/M) × N_A. Common misuse: substituting the mass in grams for N, or mixing λ in y⁻¹ with an activity in Bq. Sanity check: 1.0 g of ²²⁶Ra has N = (1.0/226) × 6.02 × 10²³ = 2.66 × 10²¹ and λ = ln2/(1600 × 3.16 × 10⁷ s) = 1.37 × 10⁻¹¹ s⁻¹, giving A ≈ 3.6 × 10¹⁰ Bq, one curie.

57. EquationHL onlyData booklet: Yes

    State the exponential decay law for activity and note the equivalent forms.

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    A = A₀e^(−λt). A = activity at time t (Bq); A₀ = initial activity (Bq); λ = decay constant (s⁻¹); t = time (s). Data-booklet status: printed. Because A = λN, the same exponential also describes the mass of the radioactive nuclide, m = m₀e^(−λt), and the corrected count rate, R = R₀e^(−λt), so any of these may be substituted directly. Conditions: background must already have been subtracted from a count rate before it is used as A. Common misuse: using an uncorrected count rate, or forgetting that the ratio A/A₀ is all that is needed, so N₀ need never be found. Sanity check: after 3 half-lives A/A₀ = e^(−3ln2) = 0.125.

58. EquationHL onlyData booklet: Yes

    State the relationship between half-life and decay constant and show where it comes from.

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    T½ = ln2/λ = 0.693/λ. T½ = half-life (s); λ = decay constant (s⁻¹); ln2 = 0.693. Data-booklet status: printed. Derivation: put N = N₀/2 in N = N₀e^(−λt) to get ½ = e^(−λT½), so ln2 = λT½. Rearranged: λ = ln2/T½. Conditions: units must match — a half-life in years gives λ in y⁻¹, which must be converted to s⁻¹ (1 y = 3.16 × 10⁷ s) before calculating an activity in Bq. Common misuse: using log₁₀2 = 0.301 instead of ln2. Sanity check: C-14 with T½ = 5730 y gives λ = 0.693/5730 = 1.21 × 10⁻⁴ y⁻¹ = 3.83 × 10⁻¹² s⁻¹.

59. EquationHL onlyData booklet: No – derive

    State the linearised form of the radioactive decay law used to find the decay constant from experimental data.

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    ln N = ln N₀ − λt, and equivalently ln A = ln A₀ − λt or ln R = ln R₀ − λt for a corrected count rate. ln N = natural logarithm of the number remaining (no unit); λ = decay constant (s⁻¹); t = time (s). Data-booklet status: not printed — derive by taking natural logs of N = N₀e^(−λt). Comparing with y = mx + c: gradient = −λ, y-intercept = ln N₀, and T½ = ln2/λ follows. Conditions: background must be subtracted first, otherwise the line curves. Common misuse: using log₁₀, which gives a gradient of −λ/2.303 (multiply by 2.303 to recover λ). Sanity check: gradient −0.0231 s⁻¹ gives T½ = 0.693/0.0231 = 30 s.

60. Graph/diagramHL only

    Describe the ln(activity) against time graph and explain how the decay constant, half-life and uncertainty are obtained from it.

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    Axes: ln(A/Bq) — or ln of the corrected count rate — on the y-axis, time t/s on the x-axis; the logarithm must be of a pure number, so state the unit divided out. Since ln A = ln A₀ − λt, the plot is a straight line of negative gradient. Gradient = −λ, so λ = −gradient and T½ = ln2/λ; the y-intercept is ln A₀, so A₀ = e^(intercept). Uncertainty: draw maximum and minimum gradient lines through the error bars and take λ_max and λ_min, then Δλ = (λ_max − λ_min)/2. A curved plot that levels off shows the background was not subtracted; a nuclide with a shorter half-life gives a steeper line with the same intercept.

61. Graph/diagramHL only

    Describe the exponential N against t decay curve and state how it changes when the decay constant or the initial number of nuclei is altered.

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    Axes: number of undecayed nuclei N (no unit) or activity A/Bq on the y-axis, time t/s on the x-axis. The curve starts at N₀, has an initial gradient of −λN₀, and decreases with a constant fractional rate, so equal time intervals always remove equal fractions — this is the graphical signature of an exponential and the reason the half-life is constant. The area under an A against t graph equals the total number of nuclei that have decayed in that interval. Doubling N₀ scales the whole curve vertically without altering its shape or half-life; doubling λ halves the half-life, making the curve fall twice as steeply while starting from the same point. The curve is asymptotic to N = 0.

62. Exam technique/trapHL only

    Explain the difference between the activity of a source and the count rate registered by a Geiger–Müller tube.

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    The trap: students treat the GM reading as the activity. Activity is the total number of decays per second in the whole source, radiated in all directions. The count rate is only what the detector actually registers, and is always much smaller. Reasons: (1) geometry — the tube subtends a small solid angle, so it intercepts only a small fraction of the emitted radiation, and that fraction falls as 1/r² for a point source; (2) efficiency — a GM tube detects nearly every α or β that enters but only about 1% of γ photons; (3) absorption and self-absorption in the source, the air gap and the tube window, which removes α almost entirely; (4) dead time at high rates. Correct approach: count rate ∝ activity for a FIXED geometry, so ratios and half-lives are still valid, but never quote a count rate in Bq. Always subtract background first.

63. Concept/explainHL onlyData booklet: Yes

    Explain the meaning of the decay constant and show how it leads to A = λN and to the exponential decay law.

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    • The decay constant λ is the probability per unit time that a given nucleus will decay, with SI unit s⁻¹ (also min⁻¹, day⁻¹, y⁻¹)
    • it is a fixed property of the isotope, unaffected by temperature, pressure or chemistry
    • for N undecayed nuclei the expected number decaying per unit time is A = λN (booklet), so activity is directly proportional to the number remaining
    • since A = −dN/dt, this gives dN/dt = −λN, whose solution is N = N₀e^(−λt) (booklet)
    • multiplying through by λ gives A = A₀e^(−λt) (booklet)
    • setting N = N₀/2 gives e^(−λT½) = ½, hence T½ = ln2/λ (booklet), so a large λ means a short half-life
    • a big λ means a very unstable, highly active isotope. Exam tip: λ is a probability per unit time, not a fraction, and it is not the wavelength.
64. Concept/explainHL onlyData booklet: Yes

    Explain how a graph of ln N (or ln A) against t is used to test for exponential decay and to find the decay constant.

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    • Start from N = N₀e^(−λt) and take natural logarithms of both sides: ln N = ln N₀ − λt
    • comparing with y = mx + c, a plot of ln N (y) against t (x) is a STRAIGHT LINE of gradient −λ and y-intercept ln N₀
    • a straight line is itself the evidence that the decay really is exponential; the raw N against t curve looks similar to several non-exponential shapes, so it cannot confirm this
    • the same works for corrected count rate: ln(C − C_background) against t has gradient −λ
    • T½ is then found from T½ = ln2/λ (booklet)
    • the uncertainty in λ comes from the maximum and minimum gradients drawn through the error bars
    • ln of a quantity with units is acceptable in IB provided the axis is labelled ln(A/Bq). Exam tip: the gradient is NEGATIVE — quote λ as its magnitude.
65. Concept/explainHL onlyData booklet: Yes

    Explain why different experimental techniques are needed to measure a very short half-life and a very long half-life.

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    • Short half-life (seconds to minutes, e.g. protactinium-234m at about 70 s): the activity falls measurably during the experiment, so record the corrected count rate at regular short intervals, plot the decay curve or ln(count rate) against t, and take λ from the gradient
    • the limits are the counting statistics in each short interval and the detector dead time
    • Long half-life (years, e.g. radium-226 at 1600 y): the activity is essentially constant over any laboratory session, so a decay curve is useless
    • instead measure the activity A with a calibrated detector of known efficiency and geometry, find the number of nuclei N from the mass of the sample using N = (m/M) × N_A, then use λ = A/N and T½ = ln2/λ
    • the main uncertainty is in the detector efficiency and in the purity of the sample.
66. Worked problemHL onlyData booklet: Yes

    A sample of iodine-131 initially contains 5.0 × 10¹⁸ undecayed nuclei. The half-life is 8.0 days. Determine the decay constant and the number of undecayed nuclei after 20 days.

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    Decay constant: λ = ln2/T½ = 0.693/8.0 = 0.0866 day⁻¹ (or 0.693/(8.0 × 86400) = 1.00 × 10⁻⁶ s⁻¹). Exponent: λt = 0.0866 × 20 = 1.733. Exponential factor: e^(−1.733) = 0.177. Nuclei remaining: N = N₀e^(−λt) = 5.0 × 10¹⁸ × 0.177 = 8.8 × 10¹⁷. Answers: λ = 0.0866 day⁻¹, N = 8.8 × 10¹⁷ (2 s.f.). Check/Trap: 20 days is 2.5 half-lives, so the fraction should be between ¼ and ⅛ — and 0.177 lies neatly between 0.25 and 0.125, confirming the arithmetic. The usual errors are keeping λ in day⁻¹ while putting t in seconds, and forgetting the minus sign in the exponent, which makes N larger than N₀.

67. Worked problemHL onlyData booklet: Yes

    Determine the activity of a 1.0 g sample of pure strontium-90, which has a half-life of 28.8 years and a molar mass of 90 g mol⁻¹.

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    Number of nuclei: N = (m/M) × N_A = (1.0/90) × 6.02 × 10²³ = 6.69 × 10²¹. Half-life in seconds: T½ = 28.8 × 365 × 24 × 3600 = 9.08 × 10⁸ s. Decay constant: λ = ln2/T½ = 0.693 / 9.08 × 10⁸ = 7.63 × 10⁻¹⁰ s⁻¹. Activity: A = λN = 7.63 × 10⁻¹⁰ × 6.69 × 10²¹ = 5.1 × 10¹² Bq. Answer: 5.1 × 10¹² Bq, i.e. about 5.1 TBq (2 s.f.). Check/Trap: λ MUST be converted to s⁻¹ before using A = λN, or the answer comes out in decays per year and is 10⁷ times too small. Also note this is the activity of the source, far larger than any count rate a GM tube would register.

68. Worked problemHL onlyData booklet: Yes

    A hospital source of iodine-131 (half-life 8.0 days) may only be used until its activity has fallen to 5.0% of the initial value. Determine how long it can be used.

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    Decay constant: λ = ln2/T½ = 0.693/8.0 = 0.0866 day⁻¹. Decay law: A = A₀e^(−λt), so A/A₀ = 0.050 = e^(−λt). Take natural logs: ln(0.050) = −λt, so −3.00 = −0.0866 t. Time: t = 3.00/0.0866 = 34.6 days ≈ 35 days (2 s.f.). Check/Trap: 35 days is about 4.3 half-lives and (½)⁴·³ ≈ 0.051, which agrees. The standard errors are taking log₁₀ instead of ln, and inverting the ratio — if you use ln(A₀/A) = +λt make sure the ratio is 20, not 0.05. Quote the answer to a sensible precision: the half-life is given to 2 s.f., so 35 days, not 34.63 days.

69. Worked problemHL onlyData booklet: Yes

    A graph of ln(A/Bq) against time t gives a straight line of gradient −0.0231 s⁻¹ and y-intercept 6.40. Determine the decay constant, the half-life and the initial activity of the source.

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    Linearised form: taking logs of A = A₀e^(−λt) gives ln A = ln A₀ − λt, so gradient = −λ and intercept = ln A₀. Decay constant: λ = 0.0231 s⁻¹ (magnitude of the gradient). Half-life: T½ = ln2/λ = 0.693/0.0231 = 30.0 s. Initial activity: ln A₀ = 6.40, so A₀ = e^(6.40) = 6.0 × 10² Bq. Answers: λ = 0.0231 s⁻¹, T½ = 30.0 s, A₀ = 6.0 × 10² Bq (3, 3 and 2 s.f.). Check/Trap: quote λ as positive — the minus sign belongs to the gradient, not to λ. The intercept must be read where t = 0 on the graph, which is only the axis crossing if the t-axis starts at zero; on a false-origin graph you must substitute a point into the equation instead.

70. Worked problemHL onlyData booklet: Yes

    The background count rate is 0.42 s⁻¹. A source gives 25.4 s⁻¹ at t = 0 and 6.67 s⁻¹ at t = 300 s. Determine the half-life and the decay constant.

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    Correct for background first: C₀ = 25.4 − 0.42 = 25.0 s⁻¹; C = 6.67 − 0.42 = 6.25 s⁻¹. Ratio: C/C₀ = 6.25/25.0 = 0.250 = ¼ = (½)², so exactly 2 half-lives have passed. Half-life: T½ = 300/2 = 150 s. Decay constant: λ = ln2/T½ = 0.693/150 = 4.62 × 10⁻³ s⁻¹. Formal route: λ = ln(C₀/C)/t = ln(4.00)/300 = 1.386/300 = 4.62 × 10⁻³ s⁻¹, giving the same result. Answers: T½ = 150 s, λ = 4.62 × 10⁻³ s⁻¹ (3 s.f.). Check/Trap: subtracting background is worth a mark on its own — using the raw values gives C/C₀ = 0.263 and a half-life about 5 s too long. Corrected count rate is proportional to activity only if the geometry is unchanged between the two readings.

71. Worked problemHL onlyData booklet: Yes

    Radon-222 has a half-life of 3.8 days. Determine the number of radon-222 nuclei, and the mass of radon, needed to give an activity of 1.0 MBq.

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    Half-life in seconds: T½ = 3.8 × 24 × 3600 = 3.28 × 10⁵ s. Decay constant: λ = ln2/T½ = 0.693 / 3.28 × 10⁵ = 2.11 × 10⁻⁶ s⁻¹. Rearrange A = λN: N = A/λ = 1.0 × 10⁶ / 2.11 × 10⁻⁶ = 4.7 × 10¹¹ nuclei. Mass: m = N × A_r × u = 4.7 × 10¹¹ × 222 × 1.66 × 10⁻²⁷ = 1.7 × 10⁻¹³ kg, i.e. about 0.17 ng. Answers: 4.7 × 10¹¹ nuclei, 1.7 × 10⁻¹³ kg (2 s.f.). Check/Trap: an activity of a million becquerels needs only a fraction of a nanogram — a useful reminder that activity is no guide to mass, and that short-lived isotopes are intensely active in tiny quantities. Do not use N = A/λ with λ in day⁻¹ and A in Bq.

72. Exam technique/trapHL onlyData booklet: Yes

    Describe how the half-life of protactinium-234m is measured in the school laboratory, and how the data are analysed.

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    Apparatus: sealed protactinium generator bottle (uranyl nitrate in acid with an organic solvent), GM tube clamped against the bottle, scaler or datalogger, stopwatch. Method: measure background for several minutes first. Shake the bottle so the protactinium transfers into the upper organic layer, allow the layers to separate, then clamp the tube beside the ORGANIC layer only and record the count in successive 10 s intervals for about 5 minutes. Repeat and average. Analysis: subtract background from every reading, plot corrected count rate against t and take several half-lives from the curve, or plot ln(corrected count rate) against t and take λ = −gradient, then T½ = ln2/λ ≈ 70 s. Controlled: geometry (never move the tube or bottle), counting interval, same sample. Limitations: short counts give large √N uncertainties; the tube must not see the aqueous layer. Improvements: use a datalogger for consistent timing, repeat and average, shield the tube from the lower layer with lead.

73. Exam technique/trapHL onlyData booklet: Yes

    Describe how the half-life of a long-lived isotope such as radium-226 is determined, since its activity does not fall measurably during an experiment.

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    The problem: over a lesson the activity of a 1600-year isotope is constant, so no decay curve can be plotted. Method: use A = λN. Measure the activity A with a detector of known efficiency and known solid angle — record the corrected count rate, then divide by the fractional solid angle (area of window / 4πr²) and by the detector efficiency to get the true activity in Bq. Find N from the sample: weigh a pure sample of mass m and use N = (m/M) × N_A. Then λ = A/N and T½ = ln2/λ. Variables controlled: source-to-detector distance, absorber-free path, same counting geometry throughout. Limitations and improvements: the efficiency and solid-angle corrections dominate the uncertainty, so calibrate against a source of known activity; sample impurity or a radioactive daughter in equilibrium inflates A; self-absorption in a thick sample lowers it, so use a thin, uniform sample.

74. Exam technique/trapHL onlyData booklet: Yes

    Explain how to obtain the uncertainty in a decay constant and half-life from a linearised ln A against t graph.

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    Method: plot ln(corrected count rate) on the y-axis against t on the x-axis. Convert the ±√N uncertainty in each count into a y-error bar: since y = ln A, the ABSOLUTE uncertainty in ln A equals the FRACTIONAL uncertainty in A, i.e. Δy = ΔA/A = 1/√N. Draw the best-fit line, then the steepest and shallowest lines that still pass through all the error bars. Then λ = |gradient of best fit| and Δλ = (max gradient − min gradient)/2. Because T½ = ln2/λ, the fractional uncertainties are equal: ΔT½/T½ = Δλ/λ, so ΔT½ = T½ × (Δλ/λ). Command-term guidance: "determine" expects the gradient method with working shown; "estimate" allows a two-point calculation. Traps: forgetting that logging converts fractional into absolute uncertainty, quoting the uncertainty to more than 1 significant figure, and rounding T½ to more figures than its uncertainty allows.

75. Exam technique/trapHL onlyData booklet: Yes

    Explain the algebraic and unit traps that cost marks in exponential decay calculations.

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    Trap 1: mismatched time units — λ from T½ in days is in day⁻¹, so t must be in days; a hospital or reactor question that quotes hours needs one consistent unit throughout. Trap 2: log base. The inverse of e^x is ln, not log₁₀; using log₁₀ scales every answer by 2.303. Trap 3: sign. In ln(A/A₀) = −λt the left side is negative because A < A₀; if you write ln(A₀/A) = +λt the ratio must be the one greater than 1. Trap 4: calculator entry — e^(−λt) must have the whole exponent bracketed. Trap 5: assuming the decay curve reaches zero; exponentials only approach the axis, so "how long until all the nuclei have decayed" has no finite answer. Command terms: "deduce" requires you to state the physics reason as well as the number; "determine" requires a full numerical route with a unit and appropriate significant figures — normally matching the least precise datum.

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